The current supplied to an air conditioner unit is 2.50 amps. The air conditioner is wired using a 10-gauge (diameter 2.588 mm) wire. The charge density is n= 8,48 x 1048 electrons/m, Find the following magnitudes. (a) current density (in A/m2) A/m2 (b) the drift velocity (in m/s) m/s

The current supplied to an air conditioner unit is 2.50 amps. The air conditioner is wired using a 10-gauge (diameter 2.588 mm) wire. The charge density is n= 8,48 x 1048 electrons/m, Find the following magnitudes. (a) current density (in A/m2) A/m2 (b) the drift velocity (in m/s) m/s

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Question

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The current supplied to an air conditioner unit is 2.50 amps. The air conditioner is wired using a 10-gauge (diameter 2.588 mm) wire. The charge density is n = 8.48 x 1028 electrons/m3, Find the
following magnitudes.
(a) current density (in A/m2)
| A/m²
(b) the drift velocity (in m/s)
m/s

Expert Solution

Introduction

The problem can be solved by using the concepts of current density and drift velocity. Current density can be defined as the current flowing per unit area (area of cross-section).

Mathematically,

$\begin{array}{rcl} Current Density & = & \frac{Current}{Area} \\ = & \frac{I}{π r^{2}} \end{array}$

Drift velocity is defined as the average velocity attained by the charged particles due to an electric field.

Mathematically it can be defined as

$v_{d} = \frac{I}{n e A}$

Here,

$I = current n = charge density e = charge of an electron A = area of cross section$

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